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26 February, 17:51

An ideally efficient heat pump delivers 1000 J of heat to room air at 300 K. If it extracted heat from 260 K outdoor air, how much of that delivered heat was originally work consumed in the transfer?

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  1. 26 February, 17:57
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    Wnet, in, = 133.33J

    Explanation:

    Given that

    Pump heat QH = 1000J

    Warm temperature TH = 300K

    Cold temperature TL = 260K

    Since the heat pump is completely reversible, the combination of coefficient of performance expression is given as,

    From first law of thermodynamics,

    COP (HP, rev) = 1 / (1-TL/TH)

    COP (HP, rev) = 1 / (1-260/300)

    COP (HP, rev) = 1 / (1-0.867)

    COP (HP, rev) = 1/0.133

    COP (HP, rev) = 7.5

    The power required to drive the the heat pump is given as

    Wnet, in = QH/COP (HP, rev)

    Wnet, in = 1000/7.5

    Wnet, in = 133.333J. QED

    So the 133.33J was the amount heat that was originally work consumed in the transfer.

    Extra ...

    According to the first law, the rate at which heat is removed from the low temperature reservoir is given as

    QL=QH-Wnet, in

    QL=1000-133.333

    QL=866.67J
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